$n_K=\dfrac{7,8}{39}=0,2(mol)$
$n_{HCl}=\dfrac{50.7,3\%}{36,5}=0,1(mol)$
$2K+2HCl\to 2KCl+H_2$
$\dfrac{0,2}{2}<\dfrac{0,1}{2}\to K$ dư, $HCl$ hết
$n_{KCl}=n_{HCl}=n_{K\text{pứ}}=0,1(mol)$
$\to n_{K\text{dư}}=0,2-0,1=0,1(mol)$
$2K+2H_2O\to 2KOH+H_2$
$\to n_{KOH}=n_{K\text{dư}}=0,1(mol)$
Ta có $n_{H_2}=\dfrac{n_K}{2}=0,1(mol)$
$\to m_{dd\text{spu}}=m_K+m_{dd HCl}-m_{H_2}= 7,8+50-0,1.2=57,6g$
$C\%_{KCl}=\dfrac{0,1.74,5.100}{57,6}=12,93\%$
$C\%_{KOH}=\dfrac{0,1.56.100}{57,6}=9,72\%$
